From 4362765128bb43080c9fc7ae4e9552a94ebd585e Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?S=C3=A9bastien=20Labb=C3=A9?= <slabqc@gmail.com>
Date: Mon, 24 Feb 2025 16:11:59 +0100
Subject: [PATCH] ajout has_claw_decomposition
---
slabbe/graph.py | 74 +++++++++++++++++++++++++++++++++++++++++++++++++
1 file changed, 74 insertions(+)
diff --git a/slabbe/graph.py b/slabbe/graph.py
index b5fcde2..e703080 100644
--- a/slabbe/graph.py
+++ b/slabbe/graph.py
@@ -1048,3 +1048,77 @@ def minimal_eulerian_paths(G, cost=None):
return paths
+
+def has_claw_decomposition(G, certificate=False):
+ r"""
+ Return whether a graph has a claw decomposition.
+
+ This is an answer to the question posted at
+ https://ask.sagemath.org/question/81610/test-if-a-graph-has-a-claw-decomposition/
+
+ INPUT:
+
+ - ``G`` -- undirected graph
+ - ``certificate`` -- boolean
+
+ OUTPUT:
+
+ a boolean or 2-tuple (boolean, solution) if certificate is True
+
+ EXAMPLES::
+
+ sage: from slabbe.graph import has_claw_decomposition
+ sage: G1 = Graph( [(0, 1), (0, 2), (0, 3), (0, 4), (1, 2), (1, 3),
+ ....: (1, 5), (2, 3), (2, 4), (3, 5), (4, 6), (4, 7), (5, 6), (5, 7), (6, 8),
+ ....: (6, 10), (7, 9), (7, 11), (8, 9), (8, 10), (9, 11), (10, 11)])
+ sage: has_claw_decomposition(G1)
+ False
+ sage: has_claw_decomposition(G1, certificate=True)
+ (False, None)
+
+ ::
+
+ sage: G2 = Graph([(0, 1), (0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4),
+ ....: (1, 5), (2, 4), (2, 5), (3, 5), (4, 5)])
+ sage: has_claw_decomposition(G2)
+ True
+ sage: has_claw_decomposition(G2, certificate=True) # random
+ (True,
+ [[(0, 1), (1, 2), (1, 5)],
+ [(0, 3), (1, 3), (3, 5)],
+ [(0, 2), (2, 4), (2, 5)],
+ [(0, 4), (1, 4), (4, 5)]])
+
+ """
+ import itertools
+ from sage.combinat.matrices.dancing_links import dlx_solver
+
+ rows = []
+
+ id_to_edge = [frozenset(edge) for edge in G.edges(labels=False)]
+ edge_to_id = {edge:i for (i,edge) in enumerate(id_to_edge)}
+
+ for u in G:
+ u_neighbors = G.neighbors(u)
+
+ for three_neighbors in itertools.combinations(u_neighbors, 3):
+ L = [edge_to_id[frozenset((u,v))] for v in three_neighbors]
+ L.sort()
+ rows.append(L)
+
+ d = dlx_solver(rows)
+
+ solution = d.one_solution()
+ has_solution = not solution is None
+
+ if not certificate:
+ return has_solution
+ else:
+ if has_solution:
+ solution_vertices = [[tuple(id_to_edge[id]) for id in rows[row_number]]
+ for row_number in solution]
+ return (has_solution, solution_vertices)
+ else:
+ return (has_solution, solution)
+
+
--
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